1.Introduction

A problem of three squares, sum of any two less third a square, i.e. is there a solution of simultaneous equation
{y^2+z^2-x^2=p^2, x^2+z^2-y^2=q^2 , x^2+y^2-z^2=r^2}. 

According to Dickson's book[1], L.Euler gave four method and parametric solution below.

(x,y,z,p,q,r)=( s(s+u)a-2t^2b, s(s+u)a+2t^2b, sta+2tb^2, sta+4t(s+u)b, sta+4t(s-u)b, sab-4t^2b), where a=3s+4u, b=s+2u.

We show simultaneous equation {y^2+z^2-x^2=p^2, x^2+z^2-y^2=q^2 , x^2+y^2-z^2=r^2} has infinitely many parametric solutions.
     
2.Theorem
     
A simultaneous equation {y^2+z^2-x^2=p^2, x^2+z^2-y^2=q^2 , x^2+y^2-z^2=r^2} has infinitely many parametric solutions.
 
Proof.

y^2+z^2-x^2=p^2................................................................(1)

x^2+z^2-y^2=q^2................................................................(2)

x^2+y^2-z^2=r^2................................................................(3)

From equation (1)+(2), (2)+(3), and (1)+(3) we obatian

2z^2 = p^2+q^2.................................................................(4)

2x^2 = q^2+r^2.................................................................(5)

2y^2 = p^2+r^2.................................................................(6)

From equation (4), we obtain a parametric solution for (z,p,q)=(1+k^2, -1+k^2-2k, 1-k^2-2k).

From equation (5), we obtain a parametric solution for (x,r)=(-(-2n^2-1+2k^2n^2+k^2-4kn^2-2k+4k^2n+4n)/(2n^2-1), -(2k^2n^2+k^2+2n^2+1-2n+2k^2n-4nk)/(2n^2-1)).

From equation (6), we obtain a quartic eqution.

Let U=n and V = 2(2n^2-1)y.

V^2 = (32k^2+64k+16k^4+16-64k^3)U^4
    + (-64k+32k^4-64k^3-32)U^3
    + (64k^2+32+32k^4)U^2
    + (-32k-32k^3+16k^4-16)U
    + 8k^2+16k+4k^4+4-16k^3....................................................(7)

Quartic equation (7) has a solution (U,V) = Q(0, 2(-1+k^2-2k)).

Hence this quartic equation (7) is birationally equivalent to an elliptic curve below.

Y^2+(8k^2+8)YX+(128+512k+384k^4+384k^2+128k^6-512k^5)Y
= X^3+(16k^4+32k^2+16)X^2+(-256+2048k^7-256k^8-2048k-5120k^2-2048k^3+6656k^4+2048k^5-5120k^6)X
-4096k^12+32768k^11-90112k^10+98304k^9-61440k^8+65536k^7+49152k^6-65536k^5-61440k^4-98304k^3-90112k^2-32768k-4096

Transformation is given, 

U = (64k^6-128k^5+64k^4-256k^3+4k^2X-64k^2-8kX-128k-64-4X)/Y

V = (-8192-192k^5X^2+1024Yk^3-1024Yk^7+1024Yk^5+2048Yk^6+335872k^10+303104k^12-384k^3X^2-2048Yk^2-1024Yk
  - 13312k^6X-192kX^2-81920k^13-96k^2X^2+2X^3k^2+8192k^14-28672Xk^7-96X^2-2X^3+20992Xk^8-180224k^9-491520k^11
  + 96X^2k^4+458752k^7-4X^3k+96X^2k^6+40960k^8-7168k^9X+1536k^10X-81920k-1536X+22528k^5X-303104k^2-491520k^3
  - 335872k^4-180224k^5-40960k^6-20992k^2X-7168kX+13312k^4X-28672k^3X)/(Y^2)
  
X = (-4V+8+16k^2+32k+4k^2V+8k^4-32k^3-8kV-32Uk^3-32Uk-16U+16Uk^4)/(U^2)

Y = (-32+16V+192Uk^4+256Uk+64kV+32k^2V-64U^2+64k^6U+192Uk^2+16k^4V+64U+64k^6U^2+64U^2k^4-64U^2k^2-256Uk^5
  - 128U^2k-128U^2k^5-256U^2k^3-192k-288k^2+128k^3+288k^4-192k^5+32k^6-64k^3V)/(U^3)


The point corresponding to point Q(0, 2(-1+k^2-2k)) is P(X,Y) = (-16k^4-32k^2-16, 512k^5-512k).

This point P is of infinite order, and the multiples kP, k = 2, 3, ...give infinitely many points.

Hence we can obtain infinitely many parametric solutions for equation (7).

Consequently, a simultaneous equation {y^2+z^2-x^2=p^2, x^2+z^2-y^2=q^2 , x^2+y^2-z^2=r^2} has infinitely many parametric solutions.

Q.E.D.


3.Examples


When k=2 we obtain a parametric solution,

x = -(1+k^2)*(k^8-8*k^6+30*k^4-8*k^2+1)
y = (-1+k^2-2*k)*(k^8+8*k^6-2*k^4+8*k^2+1)
z = (1+k^2)*(k^4-4*k-1)*(k^4-4*k^3-1)
p = (-1+k^2-2*k)*(k^4-4*k-1)*(k^4-4*k^3-1)
q = (k^2+2*k-1)*(k^4-4*k-1)*(k^4-4*k^3-1)
r = (-1+k^2-2*k)*(k^4+4*k-1)*(k^4+4*k^3-1).

k is arbitrary.

Similarly, when k=3 we obtain a parametric solution,

x = -(1+k^2)*(k^24+8*k^22-166*k^20+488*k^18-625*k^16+5648*k^14-6612*k^12+5648*k^10-625*k^8+488*k^6-166*k^4+8*k^2+1)
y = (k^24-8*k^22+122*k^20+24*k^18-753*k^16+2032*k^14+1260*k^12+2032*k^10-753*k^8+24*k^6+122*k^4-8*k^2+1)*(-1+k^2-2*k)
z = (1+k^2)*(k^12+4*k^11-16*k^10-3*k^8-8*k^7+16*k^6-64*k^5+3*k^4+4*k^3-1)*(k^12+4*k^9-3*k^8-64*k^7-16*k^6-8*k^5+3*k^4+16*k^2+4*k-1)
p = (-1+k^2-2*k)*(k^12+4*k^11-16*k^10-3*k^8-8*k^7+16*k^6-64*k^5+3*k^4+4*k^3-1)*(k^12+4*k^9-3*k^8-64*k^7-16*k^6-8*k^5+3*k^4+16*k^2+4*k-1)
q = (k^2+2*k-1)*(k^12+4*k^9-3*k^8-64*k^7-16*k^6-8*k^5+3*k^4+16*k^2+4*k-1)*(k^12+4*k^11-16*k^10-3*k^8-8*k^7+16*k^6-64*k^5+3*k^4+4*k^3-1)
r = (-1+k^2-2*k)*(k^12-4*k^9-3*k^8+64*k^7-16*k^6+8*k^5+3*k^4+16*k^2-4*k-1)*(k^12-4*k^11-16*k^10-3*k^8+8*k^7+16*k^6+64*k^5+3*k^4-4*k^3-1).

4.Reference

[1].L.E. Dickson:History of theory of numbers, vol 2.






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